Problem: Let $S$ be a surface in 3D described by the equation $z = y^{1.5} + xy - 5$. What is the equation of the plane tangent to $S$ at $(-2, 4)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $z = 5 - 4(x + 2) + (y - 4)$ (Choice B) B $z = 4(x - 2) + (y - 4)$ (Choice C) C $z = -5 + 4(x + 2) + (y - 4)$ (Choice D) D $z = -5 - 4(x + 2) - (y - 4)$
Explanation: The equation for a tangent plane of an explicitly defined surface $z = f(x, y)$ at the point $(a, b)$ is: $f(a, b) + f_x(x - a) + f_y(y - b) = z$ [What's the intuition behind the formula?] Let's find $f(-2, 4)$, $f_x$, and $f_y$. $\begin{aligned} &f(-2, 4) = 8 + (-2)(4) - 5 = -5 \\ \\ &f_x = y = 4 \\ \\ &f_y = \dfrac{3}{2}\sqrt{y} + x = 3 - 2 = 1 \end{aligned}$ Putting it all together, here's the equation for the tangent plane of $S$ at $(-2, 4)$ : $z = -5 + 4(x + 2) + (y - 4)$